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Potentiometer Loading

Both analog potentiometers and digipots have output or wiper terminals that drive load resistances. How does potentiometer loading affect output linearity? How light must the loading be to achieve a desired linearity? This article addresses these questions.

The circuit of either an analog potentiometer (or pot , for short) or a digipot is basically the same, as shown below (left figure) using an analog pot symbol. The analog pot voltage-divider attenuation, T , is varied mechanically while the digipot receives a digital input value that selects where in the pot resistance the output terminal – the digital wiper – is to be attached, using many analog switches for selection.

The wiper of the pot – the arrow in the symbol – moves across a constant resistance between the top and bottom terminals, the pot resistance value of R . Moving the wiper causes the divider to change T by apportioning R among RT and RB (center figure) with

The pot terminals between R are shown connected to voltage sources at top, VT , and bottom, VB . (If the top and bottom terminals are connected to voltages in series with resistances, add the respective resistances to RT and RB .) The pot output terminal is loaded by resistance RL . The open-circuit (RL → ∞ ) divider transfer function is

where the voltage across the pot is

The pot resistance at the wiper port is the parallel combination of the two resistances;

with maximum RW = R /4. The wiper voltage without RL by superposition is thus

As T varies, VW varies with it linearly, where T is the rate of change (slope) of VW and VB is the offset.

Now consider the case where the pot is loaded. Then the circuit can be Thevenized so that the pot wiper port is made equivalent to an open-circuit voltage source, VW , in series with the equivalent wiper resistance, RW , as shown in the figure above on the right. Solving this simple divider circuit for output voltage, VO , as a ratio of VW ,

To unclutter the design formulas, let RL be some fraction or multiple, α , the loading factor of R ;

Then substituting for RL ,

This voltage-divider transfer function is that of the loaded voltage normalized to the open-circuit pot divider voltage. Ideally, with no loading, VO = VW . As loading increases, the loaded-divider ratio decreases. As a function of T , the wiper position, it shows the extent to which VO departs from the ideal linear case of VW . Some curves of VO /VW as a function of T and loading parameter α are shown in the graph below, where

Two observations can be made of this graph. First, the curves compress to a lower than ideal value, with maximum departure at midscale, T = 1/2. Second, as loading increases, the compression flattens, with steeper sides at the extremes of the pot range, and closer to constant in the midrange. Compression error is

For α = 1 (RL = RW ), ε is about 20 %. For α = 2.5, it is about 9 % and for α = 25, is close to 1 %. The formula evaluated at maximum error, or T = 1/2, is

Then maximum compression error is

Thus, to achieve a given maximum error as expressed by compression, and in bits, the following table gives some values, where the conversion of ε to number of bits, n , of accuracy is

Then for n bits of maximum error,

The maximum error at T = 1/2 can be verified by taking the derivative of VO /VW of T , setting it equal to zero, and solving for T at which the minimum value occurs:

In conclusion, digipots with 250 positions (≈ 8 bits of resolution) require that RL be greater than 64 times the pot R value for less than one bit of error. For 10-turn analog pots with linear readouts having about 3 decades of resolution, loading should not exceed 256 x R for one bit of error. For single-turn analog pots, a loading of 25 x R will keep the error within a percent. Therefore, loading of a 10 kΩ pot must be no more than 250 kΩ for < 1 % compression error, 25 kΩ for < 10 %, and 10 kΩ for < 20 % error.

7 comments on “Potentiometer Loading

  1. GSKrasle
    June 6, 2017

    The nonlinear effects of loading the wiper of a pot can actually be USEFUL: If you add a resistor in series with the wiper, and after that a resistor to the upper terminal, and one to the lower, then by judicious choice of values, the output voltage may be made to approximate either a log or an exponential transfer function vs position, and to do it better (and obviously cheaper) than the 'log-' or 'audio-' pots that are available. For ganged pots, the tracking is also improved.

    The algebra and curve-fitting for this five-resistor circuit (two values being linked by the constitutive relation that their sum be constant) is more challenging than you'd think, and is surely the nonrecurring cost that persuades manufacturers to keep buying the worse and more-expensive products, a recurring cost and an admission of inadequate Engineering. 

    I have posted my math online before, both the algebraic solution of the circuit and the values for curve-fits, but I would love to see other derivations, if just to confirm my own.

     

    I also hope you will address the power-rating issues with pots: ones in lab-drawers often come to have have the ends of their ranges ruined by overcurrent. And also the practice of not leaving the third terminal of a variable-resistor open, so as to reduce noise due to a dirty/'iffy'” wiper contact.

  2. antedeluvian
    June 7, 2017

    @GSKrasle

    If you add a resistor in series with the wiper, and after that a resistor to the upper terminal, and one to the lower, 


    Ilearned about this technique at a Maxim seminar as a method to combat thermal effects. I described it in my blog on potentiometers a while ago.

    The algebra and curve-fitting for this five-resistor circuit (two values being linked by the constitutive relation that their sum be constant) is more challenging than you'd think


    I tried several times to develop a model for this and failed. I even planned on using it as fodder for one of my ideas on the use of Excel in Electronics, but my skills and patience were insufficient.

    I have posted my math online before

    I would really like to see this. Could you provide a pointer, please?

     

  3. GSKrasle
    June 9, 2017

    Aubrey,

    It's somewhere on EETimes or Planet Analog, I'm sure, but hete's a synopsis:

    I enjoy algebra, so I tried the solution. I'm not CERTAIN I'm correct, (because nobody cares to check me), but the solution is usable.

    I used iterations of the solver in Excel2007 to least-squares curve-fit to simple exponantial and log curves.

    The “meat” is the algebraic expression in col G for Vo/Vi:
    =((1/B16)+(1/D16)-(((C16+B16)*((C16*F16)/(C16+F16)))/(B16*C16*(B16+((C16*F16)/(C16+F16))))))/((1/D16)+(1/E16)+(((C16+B16)*((B16*C16)/(B16+C16)))/(B16*C16*(F16+((B16*C16)/(B16+C16))))))

    A: pot position index B: upper pot R (to Vi), C: lower one (to GND), D: R from Vi to Vo, E: R from Vo to GND, F: R in wiper to Vo, G: Vo/Vi. The rest calculates the target curve and the squared difference; I used to solver to minimize the summed squared difference by futzing the values of the other three resistors.

    If someone with better editorial access could intervene, I'll try to add some images:

    Here are the optimal fits to the example curves:

     

    Here is the top of the table I ended-up with (however badly it formats here:

    nicr: R1+R3: 1.0E+3Ω                 SoS: AdjFact:
    0.01 1.0E+3                 0.077669 41.5043
    n RP1: RP2: R1: 1.0E+9Ω R2: 943.5E+0Ω R3: 000.0E+0Ω Vo/Vi Exp adjExp Delt^2 Delt Delt^2, Adj
    0 1000 0 1.0E+9 943.5E+0 000.0E+0 #DIV/0! 0 0     #VALUE!
    0.01 990 10 1.0E+9 943.5E+0 000.0E+0 0.009896 0.01005 0.005849 1.64E-05 0.004047 0.00068
    0.02 980 20 1.0E+9 943.5E+0 000.0E+0 0.019593 0.020201 0.011757 6.14E-05 0.007836 0.002549
  4. GSKrasle
    June 10, 2017

    Ugh: I wish I could edit that!

    If you can privately e-mail me, I can give you the spreadsheet and plots. I SHOULD revisit it and make it prettier, but it serves. 

    I should also mention that I have seen the curves for the commercial 'log' (actually 'exponential') pots, and they are poor curve-fits, certainly worse than my solution.

    Try THIS though!
    https://xkcd.com/356/
    I worked a related problem in three dimensions, and the answer is 376.73.

    I still haven't worked-out its relation to ^(?=(?!(.)1)([^DO:105-93+30])(?-1)(?< !d(?<=(?![5-90-3])d))).[^WHY?]$

     

  5. antedeluvian
    June 11, 2017

    @GSKrasle I am travelling at the moment and so I haven't even had a chance to consider you previous post. I don't have you email, but since mine is fairly well publicized, we could start with that

    a k a g a n  a t  e m p h a t e c  d o t  c o m

    – remove all the spaces and replace at and dot with their symbols

    -Aubrey

  6. Mark M
    June 11, 2017

    It is a very interesting subject

    Thanks

    Mark

  7. D Feucht
    June 12, 2017

    GS & Aubrey,

    Thanks for your substantial additions as comments to my article. GS, perhaps you might do a sequel article to this one and put in your algebra. I submit articles to Steve Taranovich as MS Word files in which the math is done with the MS Word math editor. All comes out okay this way.

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